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# 统计代写|回归分析代写Regression Analysis代考|The Causal Model

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## 统计代写|回归分析代写Regression Analysis代考|The Causal Model

If you want the coefficient of $X$ to be a causal effect, you need a different model than the conditional mean model, one that includes many more $X^{\prime}$. Here is one:
$$Y=\gamma_0+\gamma_1 X_1+\gamma_2 X_2+\ldots+\gamma_k X_k+\varepsilon^{\prime}$$
In this model:

$X_1$ is your main $X$ variable of interest, the one whose causal effect on $Y$ you wish to measure.

$X_2$ through $X_k$ are all other (usually unmeasured) variables that also causally affect $Y$. In this model, changes (manipulations) in $X_1$ cause changes in the distribution of $Y$ when all other possible causal variables $X_2-X_k$ are held fixed.

$\varepsilon^{\prime}$ is a random error term. This term might be identically zero, in which case the causal model is a deterministic model, and this does not change any of the arguments below. Otherwise, with enough $X^{\prime}$ s, it is reasonable to assume that this term is uncorrelated with everything; e.g., $\varepsilon^{\prime}$ might be subatomic quantum noise.
You can re-arrange the causal model as follows:
$$Y=\gamma_0+\gamma_1 X+\delta$$
where $X=X_1$ and
$$\delta=\gamma_2 X_2+\ldots+\gamma_k X_k+\varepsilon^{\prime}$$
In this model, $\operatorname{Cov}(X, \delta) \neq 0$. Instead, $\operatorname{Cov}(X, \delta)=\sum_{j=2}^k \gamma_j \operatorname{Cov}\left(X, X_j\right)$. Thus, applying Theorem 6.5, the OLS estimate $\hat{\sigma}{x y} / \hat{\sigma}_x^2$ is inconsistent for $\gamma_1$, with probability limit $\gamma_1+\sum{j=2}^k \gamma_j \operatorname{Cov}\left(X, X_j\right) / \sigma_x^2$.

## 统计代写|回归分析代写Regression Analysis代考|The Instrumental Variable Method

The goal is to come up with an estimator whose probability limit is $\gamma_1$. If you could measure all the relevant unobserved confounders $X_2, \ldots, X_k$, then the simple OLS multiple regression estimate of $\gamma_1$ in model (1) would do the trick. But you usually cannot. And even if you could, there might be hundreds of such variables, and you would not want to run OLS with so many predictors. What to do? Try to find an instrumental variable.

Consider the model $Y=\gamma_0+\gamma_1 X+\delta$, where $\gamma_1$ is the causal effect of $X$. An instrumental variable is a variable $Z$ such that:

1. $Z$ is correlated with $X$ (preferably reasonably strongly correlated), and
2. $Z$ is uncorrelated with $\delta$

The instrumental variable (IV) estimator of $\gamma_1$
The instrumental variable estimator of $\gamma_1$ is given by
$$\hat{\gamma}1=\frac{\hat{\sigma}{z y}}{\hat{\sigma}{z x}}$$ Theorem 6.6: Consistency of the IV Estimator Assume the data pairs $\left(X_i, Y_i, Z_i\right)$ are sampled iid from $p(x, y, z)$, with all variances finite. Assume in addition that $Z$ is an instrumental variable to the causal model $Y=\gamma_0+\gamma_1 X+\delta$. Then $\hat{\gamma}_1=\hat{\sigma}{z y} / \hat{\sigma}_{z x}$ is a consistent estimator of $\gamma_1$.

## 计代写|回归分析代写Regression Analysis代考|The Causal Model

$$Y=\gamma_0+\gamma_1 X_1+\gamma_2 X_2+\ldots+\gamma_k X_k+\varepsilon^{\prime}$$

$X_1$ 是您感兴趣的主要$X$变量，您希望测量其对$Y$的因果影响。

$X_2$ 通过$X_k$，所有其他(通常未测量的)变量也会对$Y$产生因果影响。在这个模型中，当所有其他可能的因果变量$X_2-X_k$保持固定时，$X_1$的变化(操纵)导致$Y$分布的变化。

$\varepsilon^{\prime}$ 是随机误差项。这一项可能等于零，在这种情况下，因果模型是确定性模型，这不会改变下面的任何论点。否则，只要有足够的$X^{\prime}$ s，就可以合理地假设这一项与一切都不相关;例如，$\varepsilon^{\prime}$可能是亚原子量子噪声。

$$Y=\gamma_0+\gamma_1 X+\delta$$

$$\delta=\gamma_2 X_2+\ldots+\gamma_k X_k+\varepsilon^{\prime}$$

## 统计代写|回归分析代写Regression Analysis代考|The Instrumental Variable Method

$Z$ 与$X$相关(最好是合理的强相关)，并且

$Z$ 与 $\delta$

$\gamma_1$的工具变量(IV)估计量
$\gamma_1$的工具变量估计量由
$$\hat{\gamma}1=\frac{\hat{\sigma}{z y}}{\hat{\sigma}{z x}}$$定理6.6:IV估计量的一致性假设数据对$\left(X_i, Y_i, Z_i\right)$从$p(x, y, z)$中抽样iid，所有方差都是有限的。另外假设$Z$是因果模型$Y=\gamma_0+\gamma_1 X+\delta$的工具变量。那么$\hat{\gamma}1=\hat{\sigma}{z y} / \hat{\sigma}{z x}$是$\gamma_1$的一致估计量。